Question

differentiate (x^3cosx-2^xtanx)​

Answer 1

Answer:

$\bold\red{ \frac{dy}{dx} = - {x}^{3} \sin(x) + 3 {x}^{2} \cos(x) - {2}^{x} { \sec }^{2} x - {2}^{x} ln(x) \tan(x) }$

Step-by-step explanation:

$y = {x}^{3} \cos(x) - {2}^{x} \tan(x) \\ = > \frac{dy}{dx} = {x}^{3} \frac{d}{dx} ( \cos \: x) + \cos(x) \frac{d}{dx} ( {x}^{3} ) - {2}^{x} \frac{d}{dx}( \tan \: x) - \tan(x ) \frac{d}{dx} ( {2}^{x}) \\ \\ = > \frac{dy}{dx} = - {x}^{3} \sin(x) + 3 {x}^{2} \cos(x) - {2}^{x} { \sec }^{2} x - {2}^{x} ln(2) \tan(x)$

Here,

concepts used are :-

$1. \: \: \frac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1} \: \\ \\ 2. \: \: \frac{d}{dx} ( \cos \: x) = - \sin \: x \\ \\ 3. \: \: \frac{d}{dx} ( \tan \: x) = { \sec }^{2} x \\ \\ 4. \frac{d}{dx} ( {a}^{x} ) = {a}^{x} ln(a )$